Java 逻辑运算符和/或优先级

Question

So everywhere I look, it says that && is evaluated first, then || is evaluated second. So either I am doing something wrong or it is wrong. Here's the code:

static boolean foo(boolean b, int id){ System.out.println(id); return b;}

static{ System.out.println(foo(true, 3) && foo(true, 1) || foo(false, 2)) }
//returns 3 1

static{ System.out.println(foo(true, 2) || foo(true, 3) && foo(true, 1)} 
//returns 2

In the first static block, && goes first, short circuits and ignores the || but in the second static block which is simply the reverse, || goes first and ignored the &&. This demonstrates left to right but according to the java doc, && has higher precedence which means && should always go first.

Some documents about precedence (logical and is higher than or ):
1. https://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html
2. https://chortle.ccsu.edu/Java5/Notes/chap40/ch40_16.html
3. http://www.cs.bilkent.edu.tr/~guvenir/courses/CS101/op_precedence.html
...

Answer 1

Higher predecence of && just means that

foo(true, 2) || foo(true, 3) && foo(true, 1)

is the same as

foo(true, 2) || (foo(true, 3) && foo(true, 1))

but not

(foo(true, 2) || foo(true, 3)) && foo(true, 1)

Nothing else. It doesn't imply anything about evaluation order.

Now, for most operators, evaluating x op y requires evaluating both x and y . If || was one of them, it would be the same as ( return is from evaluating || , not from the entire method)

boolean tmp1 = foo(true, 2);
boolean tmp2 = foo(true, 3) && foo(true, 1);
return tmp1 || tmp2;

and && would indeed "go first". But there are three operators which don't work like that: && , || and ?: . Instead you get

boolean tmp1 = foo(true, 2);
if (tmp1) {
    return true; 
} else {
    return foo(true, 3) && foo(true, 1);
}